Question

SELF 5.32 The quality control manager of Marilyn's Cook-

Test ies is inspecting a batch of chocolate-chip cookies that
has just been baked. If the production process is in control, the
mean number of chocolate-chip parts per cookie is 6.0. What is the
probability that in any particular cookie being inspected
a. fewer than five chocolate-chip parts will be found?
b. exactly five chocolate-chip parts will be found?
c. five or more chocolate-chip parts will be found?
d. either four or five chocolate-chip parts will be found?
5.33 Refer to Problem 5.32. How many cookies in a batch of 100
should the manager expect to discard if company policy requires
that all chocolate-chip cookies sold have at least four chocolate-
chip parts?

Answer 1

Answer: This question is most likely assuming a poisson distribution; it is impossible to compute without some distribution clarified, as otherwise we don't have applicable information.

Poison distribution is:

(λke-k) / k!

Where k is the number of outcomes you are looking for, and λ is the accepted average number of outcomes. So, λ is 6 for this example, and we are looking for values of k to use.

Solving problem b is simply substituting the value.

Prob of 5 parts: (65e-5) / 5!, using a calculator and working it out, this approximates to 0.437.

Solving problem a can be used for c; to get a, do the above equation 5 times for k = 0, 1, 2, 3 and 4, and add them all together. Though somewhat tedious, this would lead to the solution. To get part c, you would use the property of complements:

The cumulative probability here would be 1 - (total prob of all scenarios with k = 0, 1, 2, 3, 4, 5). You can get that total probability by adding the answers to part a and b together.

Finally, regarding part d, the above work can be redone with a new formula, now (λke-k) / k! would have λ as 5. This would reduce the answer for a, but increase it for b and c, which you can verify by running the calculations.

Explanation: